$\lnot (A\iff B)$ is logically equivalent to $A\iff \lnot B$ 
Show $\lnot (A\iff B)$ is logically equivalent to $A\iff \lnot B$.

Making some calculations I got this 
$$\lnot (A\iff B)=(A\land\lnot B)\lor(B\land\lnot A) \tag{1}$$ and
$$(A\iff \lnot B)=(\lnot A\lor\lnot B)\land(\lnot B\lor\lnot A)\tag{2}$$
I don't know how to make a relation between (1) and (2). I believe there is a trick in the solution.
Could someone lighten my brain?
 A: Your calculations are not correct. Recall that $(A\implies B)=\lnot A\lor  B$.
Now note that for the distributive law,
$$(A\iff \lnot B)=(\lnot A\lor\lnot B)\land(B\lor A)=
(\lnot A\land B)\lor( \lnot A\land  A)\lor (\lnot B\land B)\lor(\lnot B\land A).
$$
Is the above expression equivalent to 
$$\lnot (A\iff B)=(A\land\lnot B)\lor(B\land\lnot A) \quad ?$$
A: Logically $A\Leftrightarrow B$ means $A$ and $B$ have equal truth values (either both true or both false). If this is not the case, they have different truth values, which then means $A$ and $\neg B$ have the same truth value. This is a consequence of the two-valuedness of classical logic. You can also use a truth table to see this more clearly.
\begin{align}
\begin{array}{cll}
A\Leftrightarrow B & 0 &1\\
0 & 1 & 0\\
1 & 0 & 1
\end{array}\qquad
\begin{array}{cll}
A\Leftrightarrow\neg B & 0 &1\\
0 & 0 & 1\\
1 & 1 & 0
\end{array}
\end{align}
From the above truth tables, the truth value of $\neg(A\Leftrightarrow B)$ is always the same as that of $A\Leftrightarrow\neg B$.
A: By logic table the first tells $A$ and $B$ have not the same logic values, the second that they have different logic values (so they are equivalent under tertium non datur assumption). I know this is not the searched answer, but was funny and worth to note.
